Question

If \( abc = 1 \) where \( a, b, c > 0 \) show that
\[ a^2 + b^2+c^2 \ge a+b+c \]
No Lagrange multipliers allowed.

Answer 1

I think it involve multiplying a side by \(abc\)

Answer 2

i just solved it after posting .. it was clever!! I'll keep the question open for a while.

Answer 3

@wio would you like a hint .. since i had it.

Answer 4

What's the hint?

Answer 5

it's AM-GM inequality. Perhaps you are familiar.
\[ 3(a^2+b^2+c^2) \ge (a+b+c)^2 \ge 3(ab+bc+ca) \]
This is half of it, the other one, i presume you know it.

Answer 6

sorry, also i forgot to specify that \( a, b, c > \) at the beginning.

Answer 7

They're greater than 0? cause that changes a lot

Answer 8

Well, the problem is stated as
"Let a; b; c be positive real numbers such that abc = 1. Prove that ... "

Answer 9

Ah ok.

Answer 10

I was messing around with algebra getting dead ends like \[
\begin{array}{rcl}
a^2 + b^2+c^2 &\ge& a+b+c \\
a^2 + b^2+c^2 &\ge& a^2bc+b^2ac+c^2ab \\
0 &\ge& (bc-1)+(ac-1)+(ab-1) \\
0 &\ge& bc+ac+ab-3 \\
3 &\ge& bc+ac+ab \\

\end{array}
\]

Answer 11

Here's a short proof hint ... i had written earlier. \[ \begin{align*} 3(a^2+b^2+c^2) & = a^2 + b^2 + c^2 + (a^2+b^2) + (b^2 + c^2) + (c^2+a^2) \\ & \ge a^2 + b^2 +c^2 + 2 a b + 2 b c + 2 c a \\ & = (a+b+c)^2 \\ & = \frac{2(a^2 + b^2 +c^2)}{2} + 2 a b + 2 b c + 2 ca \\ & \ge \frac{ 2 a b + 2 b c + 2 ca }{2} + 2 a b + 2 b c + 2 ca \\ & = 3(a+b+c) \end{align*} \]

Answer 12

the other hint would to be make use of AM-GM inequality for 3 variables.

Answer 13

\[ 3(a^2+b^2+c^2) \ge (a+b+c)^2 \\
3 = 3 \sqrt[3]{abc} \le (a+b+c)
\]
Dividing will do the job!!

Answer 14

(a+b+c)/3 >= (abc)^1/3
a+b+c>=3
Now,
a^2+b^2+c^2=(a+b+c)^2 -2ab -2bc-ca
SO,
a^2+b^2+c^2>=(a+b+c)^2 .....i
Since, a+b+c>=3
(a+b+c)^2 >=a+b+c ....ii
Thus,
from i and ii,
a^2+b^2+c^2 >=a+b+c